Question

Integrals of the form $\int R(x,\sqrt{a{x}^2+bx+c})dx$ are calculated with the aid of one of the three Euler substitutions.
$\text{I.}\sqrt{a{x}^2+bx+c}=t\pm x\sqrt{a}$ if $a>0;$
$\text{II.}\sqrt{a{x}^2+bx+c}=tx\pm \sqrt{c}$ if $c>0;$
$\text{III.}\sqrt{a{x}^2+bx+c}=(x-a)t\pm x$ if $a{x}^2+bx+c=a(x-\alpha )(x-\beta )$ i.e., if $\alpha$ is a real root of $a{x}^2+bx+c=0.$

Which of the following functions does not appear in the primitive of $\frac{1}{1+\sqrt{x^2+2x+2}}$ if $t$ is a function of $x?$

• A. ${\log }_e|t+1|$
• B. ${\log }_e|t+2|$
• C. $\frac{1}{t+2}$
• D. none of these

Answer 1

The correct option is D none of these

Here, $a=1>0$
Therefore, we make the substitution $\sqrt{x^2+2x+2}=t-x.$
Squaring both sides of this equality and reducing the similar terms, we get
$2x+2tx=t^2-2$
$\Rightarrow x=\frac{t^2-2}{2(1+t)}$
$\Rightarrow dx=\frac{t^2+2t+2}{2(1+t{)}^2}dt$

$1+\sqrt{x^2+2x+2}=1+t-\frac{t^2-2}{2(1+t)}$
$=\frac{t^2+4t+4}{2(1+t)}$

Substituting into the integral, we get
$I=\int \frac{2(1+t)(t^2+2t+2)}{(t^2+4t+4)2(1+t{)}^2}dt$
$=\int \frac{(t^2+2t+2)}{(1+t)(t+2{)}^2}dt$

Now, let us expand the obtained proper rational fraction into partial fractions:
$\frac{t^2+2t+2}{(t+1)(t+2{)}^2}=\frac{A}{t+1}+\frac{B}{t+2}+\frac{C}{(t+2{)}^2}$

Integrating both the sides,
$\int \frac{t^2+2t+2}{(t+1)(t+2{)}^2}dt=\int (\frac{A}{t+1}+\frac{B}{t+2}+\frac{C}{(t+2{)}^2})dt$
$=A{\log }_e|t+1|+B{\log }_e|t+2|-C\frac{1}{t+2}+c$